13问答网 > (3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)

(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)

麻烦用简便算法，谢谢

2025-02-27 03:27:35

推荐回答（5个）

回答1：

原式乘以(3-1)再除以2
原式=(3^64-1)/2=1716841910146256242328924544640

回答2：

(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
＝（3-1）(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)÷2
＝（3^2-1）(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)÷2
＝（3^4-1）(3^4+1)(3^8+1)(3^16+1)(3^32+1)÷2
=(3^8-1)(3^8+1)(3^16+1)(3^32+1)÷2
=(3^16-1)(3^16+1)(3^32+1)÷2
=(3^32-1)(3^32+1)÷2
=(3^64-1)/2

回答3：

(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)/(3-1)
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)/2
=(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)/2
……
=(3^64-1)/2

回答4：

(3-1)*...

(3^64-1)/2

回答5：

(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=(3-1)(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)/2
=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)/2
=……
＝(3^64-1)/2